第一道矩阵快速幂的题;模板题;
#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm> using namespace std; #define INF 0x3f3f3f3f
typedef long long ll;
const ll maxn = ;
int n;
int m; ll euler[maxn];
ll fun_[maxn]; struct node
{
int a[][];
node()
{
memset(a,,sizeof(a));
}
node operator *(const node& m2) const
{
node m1;
for (int i=; i<n; i++)
{
for (int j=; j<n; j++)
{
m1.a[i][j]=;
for (int k=; k<n; k++)
{
m1.a[i][j]+=a[i][k]*m2.a[k][j];
m1.a[i][j]%=;
}
}
}
return m1;
}
node operator +(const node& m2) const
{
node m1;
for (int i=; i<n; i++)
{
for (int j=; j<n; j++)
{
m1.a[i][j]=a[i][j]+m2.a[i][j];
m1.a[i][j]%=;
}
}
return m1;
} } A,E; void init()
{
memset(E.a,,sizeof(E.a));
for (int i=;i<n;i++)
{
// for (int j=0;j<n;j++)
// {
// if (i==j)
E.a[i][i]=;
// }
}
} node quick_mod_(node c,int k)
{
node e=E;
while (k)
{
if(k&) e=e*c;
c = c*c;
k>>=;
}
return e;
} node find_(int k)
{
if (k==)
{
return A;
}
if(k%==)
{
k/=;
node nod=quick_mod_(A,k)+E;
nod=nod*find_(k);
return nod;
}
else
{
node t=quick_mod_(A,k);
k--;
k/=;
node nod=quick_mod_(A,k)+E;
nod=nod*find_(k);
//cout<<nod.a[0][3]<<endl;
nod = nod +t;
return nod;
}
} void print_(node p)
{
for (int i=;i<n;i++)
{
for (int j=;j<n;j++)
{
if (!j)
cout<<p.a[i][j];
else
cout<<" "<<p.a[i][j];
}cout<<endl;
}
} int main()
{
int kase=; while (cin>>n>>m&&(m+n))
{
init();
if (kase) cout<<endl;
for (int i=; i<n; i++)
{
for (int j=; j<n; j++)
{
cin>>A.a[i][j];
A.a[i][j]%=;
}
}
node q=find_(m);
// cout<<endl;
// cout<<q.a[0][0]<<" "<<q.a[0][1]<<" "<<q.a[0][2]<<endl<<endl;
print_(q);
kase++;
}
return ;
} /*
3 2
0 2 0
0 0 2
0 0 0 */
倍增法的 解释: